Show that a ring is commutative if it has the property that ab = ca implies b = c when $a\neq 0$

Solution 1:

I think the claim is fine and your attempt at a proof is in the right direction. If you have some ring $R$ in which $\forall a,b,c\in R (ab=ca\wedge a\neq 0)\implies b=c$, then $R$ is commutative.

Your proof works just fine, with a couple of slight modifications. First we show that if $x,y\in R$ and $x\neq 0$, $y\neq 0$ then $xy=yx$. We do this exactly the way you did. We let $a=x$, $b=yx$ and $c=xy$ then $ab=x(yx)=(xy)x=ca$ where the first and last $=$ are just using the definitions and the middle $=$ follows from associativity of multiplication. Now we can use the property since $ab=ca$ and $a\neq 0$ we get $b=c$ which means $xy=yx$.

Dealing with $x$ or $y$ equal to $0$ is easy since you get $x0=0=0x$.

The reason that the statement uses $a\neq 0$ is most probably so that you get any rings that actually satisfy the claim. Since obviously for $a=0$ if a ring were to satisfy the claim you would get that all elements must be equal.